I figured an equation that works for this series but have no idea how to calculate it once I have the sum...none of the rules in our book seem to apply.
Try to use some algebra
$\displaystyle \left( \frac{1}{2}\right)^{2n-5}=\left( \frac{1}{2}\right)^{2(n-1)-3}=\left( \frac{1}{4}\right)^{n-1} \left( \frac{1}{2}\right)^{-3} =8\left( \frac{1}{4}\right)^{n-1}$
So your sum is the same as
$\displaystyle 8\sum_{n=1}^{\infty}\left( \frac{3}{4} \right)^{n-1} $
So $\displaystyle r=\frac{3}{4}$
So the sum of the series $\displaystyle 8\cdot\frac{1}{1-\frac{3}{4}}=8\cdot\frac{1}{\frac{1}{4}}=32$
P.S. I think your index starts at n=1 not at n =0.
I find it easier to look at these like this:
$\displaystyle a_0 = 8$
$\displaystyle a_n = \frac 34a_{n-1}$
Looking at that a little bit (you can do the first few a_n if it helps to see, then you will be familiar enough to spot it right off) it should be clear that $\displaystyle a_n = 8*\left(\frac 34\right)^n$
And since you want to sum all of them, you get
$\displaystyle \sum_{n=0}^\infty a_n$
$\displaystyle =\sum_{n=0}^\infty 8*\left(\frac 34\right)^n$
$\displaystyle =8\sum_{n=0}^\infty \left(\frac 34\right)^n$
At which point you have the same geometric series that TheEmptySet and Soroban solved for.